# Induction well ordering principle [duplicate]

Can someone help me with the following question. I have mangaged to solve this question using well ordering prinicple but cant proof it by the induction method. I cant proof that n+1 holds in the equation. Please be as detailed as possible. The question could be found in link below, it is listed as question number 2.

Prove by either the Well Ordering Principle or induction that for all nonnegative integers $n$: $$\sum_{i=0}^n i^3 = \left(\frac{n(n+1)}2\right)^2.\tag1$$

## marked as duplicate by MJD, N. F. Taussig, Asaf Karagila♦, Jeremy Rickard, drhabJun 25 '15 at 14:47

See, for $n=0,1$, statement is true.Now assume that statement is true for $n=k$ i.e. $\sum_{i=0}^k i^3 = \left(\frac{k(k+1)}2\right)^2.\tag1$, then you have to show for $n=k+1$ statement also holds. $$\sum_{i=0}^{k+1} i^3 = \sum_{i=0}^k i^3+(k+1)^3$$. Now using equation (1) we have, \begin{align*}\left(\frac{k(k+1)}2\right)^2+(k+1)^3 &= \frac{k^2(k+1)^2}{4} + (k+1)^3\\ &= \frac{k^2(k+1)^2 + 4(k+1)^3}{4}\\ &= \frac{(k+1)^2(k^2 + 4k + 4)}{4}\\ &= \frac{(k+1)^2(k+2)^2}{4}.\\ &= \frac{(k+1)^2((k+1)+1)^2}{4} &=\left(\frac{(k+1)((k+1)+1)}2\right)^2 \end{align*}

Hence, statement holds for $n=k+1$. Then by Principle of induction statement is true for all $n$

Hint:

Let $F(n)= \sum_{i=0}^n i^3$ and use : $F(n)-F(n-1)=n^3$ (here is used induction).

This gives: $$\left(\dfrac{n(n+1)}{2}\right)^2-\left(\dfrac{(n-1)n}{2}\right)^2=n^3$$ That you can easely verify.

Proof by Induction:

Base Case: For n = 1, LHS = 1, RHS = $(\frac{1*2}{2})^2$ = 1

Induction Hypothesis: For some n, assume $\sum\limits_{i=1}^n i^3 = (\frac{n(n+1)}{2})^2$

For n+1, $\sum\limits_{i=1}^{n+1} i^3 = (\frac{n(n+1)}{2})^2 + (n+1)^3$

$= (n+1)^2 (\frac{n^2 + 4n + 4}{4})$ $= (\frac{(n+1)(n+2)}{2})^2$